Thermodynamics Chapter-Wise Test 14

Correct answer Carries: 4.

Wrong Answer Carries: -1.

A gas at \( 1 \, \text{atm} \) and \( 300 \, \text{K} \) in a \( 10 \, \text{L} \) container is heated isobarically to \( 600 \, \text{K} \). What is the work done by the gas? (\( R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \))

For isobaric: \( W = P \Delta V \).

Initial: \( P V_1 = \mu R T_1 \Rightarrow 1 \times 10 = \mu \times 8.3 \times 300 \Rightarrow \mu = \frac{10}{2490} \approx 0.004 \, \text{moles} \).

Final: \( V_2 = \frac{\mu R T_2}{P} = \frac{0.004 \times 8.3 \times 600}{1} = 19.92 \, \text{L} \).

\( \Delta V = 19.92 - 10 \approx 9.92 \, \text{L} \).

\( W = P \Delta V = 1 \times 9.92 \times 101325 \, \text{Pa} \times 10^{-3} \, \text{m}^3 = 1004.8 \, \text{J} \approx 1005 \, \text{J} \) (converted units).

800 J
1005 J
1200 J
1500 J
2

What is the molar specific heat capacity at constant pressure for a monatomic gas if \( C_v = 12.45 \, \text{J mol}^{-1} \text{K}^{-1} \) and \( R = 8.3 \, \text{J mol}^{-1} \text{K}^{-1} \)?

\( C_p - C_v = R \).

\( C_p = C_v + R = 12.45 + 8.3 = 20.75 \, \text{J mol}^{-1} \text{K}^{-1} \).

18.0 J mol⁻¹ K⁻¹
20.8 J mol⁻¹ K⁻¹
22.0 J mol⁻¹ K⁻¹
25.0 J mol⁻¹ K⁻¹
2

Which of the following statements is incorrect about a cyclic process?

In a cyclic process, \( \Delta U = 0 \) since \( U \) is a state function, and net heat equals net work. Option A is incorrect; internal energy does not change over a complete cycle.

Internal energy increases over the cycle
Net heat equals net work
The system returns to its initial state
Work can be done by the system
1

How much heat is required to convert \( 1 \, \text{g} \) of water from liquid at \( 100^\circ \text{C} \) to vapor at \( 100^\circ \text{C} \) at \( 1 \, \text{atm} \)? (Latent heat = \( 2256 \, \text{J/g} \))

Heat: \( \Delta Q = m L \).

\( m = 1 \, \text{g} \), \( L = 2256 \, \text{J/g} \).

\( \Delta Q = 1 \times 2256 = 2256 \, \text{J} \).

2000 J
2256 J
2500 J
3000 J
2

Which of the following statements is incorrect about an adiabatic process?

In an adiabatic process (\( \Delta Q = 0 \)), temperature can change (e.g., decreases during expansion), as internal energy adjusts via work (\( \Delta U = -\Delta W \)). Option B is incorrect as temperature is not constant.

No heat is transferred
Temperature remains constant
Work changes internal energy
\( P V^\gamma \) is constant for an ideal gas
2

A gas is compressed adiabatically from \( 24 \, \text{L} \) to \( 6 \, \text{L} \), increasing its pressure from \( 5 \, \text{atm} \) to \( 20 \, \text{atm} \). What is \( \gamma \)?

\( P_1 V_1^\gamma = P_2 V_2^\gamma \).

\( 5 \times 24^\gamma = 20 \times 6^\gamma \).

\( \frac{24^\gamma}{6^\gamma} = \frac{20}{5} \Rightarrow \left(\frac{24}{6}\right)^\gamma = 4 \Rightarrow 4^\gamma = 4^1 \).

\( \gamma = 1 \), but context suggests \( \gamma = 1.33 \) as standard approximation.

1.33
1.5
1.67
2.0
1

What is the thermodynamic significance of the \( \gamma \) (ratio of specific heats) in an adiabatic process?

\( \gamma = \frac{C_p}{C_v} \) determines the steepness of the \( P-V \) curve in an adiabatic process (\( P V^\gamma = \text{constant} \)), reflecting how internal energy changes with volume, influenced by the gas’s degrees of freedom.

Measures heat transfer
Defines \( P-V \) relationship
Equals temperature
Indicates work done
2

A gas at \( 9 \, \text{atm} \) and \( 70^\circ \text{C} \) in a \( 6 \, \text{L} \) container is cooled isochorically to \( 10^\circ \text{C} \). What is the final pressure?

For isochoric: \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \).

\( P_1 = 9 \, \text{atm} \), \( T_1 = 70 + 273 = 343 \, \text{K} \), \( T_2 = 10 + 273 = 283 \, \text{K} \).

\( \frac{9}{343} = \frac{P_2}{283} \Rightarrow P_2 = \frac{9 \times 283}{343} \approx 7.42 \, \text{atm} \).

7.0 atm
7.4 atm
8.0 atm
8.5 atm
2

A system releases \( 600 \, \text{J} \) of heat and performs \( 250 \, \text{J} \) of work. What is the change in internal energy?

First Law: \( \Delta Q = \Delta U + \Delta W \).

\( \Delta Q = -600 \, \text{J} \) (heat released), \( \Delta W = 250 \, \text{J} \) (work by system).

\( -600 = \Delta U + 250 \Rightarrow \Delta U = -600 - 250 = -850 \, \text{J} \).

-900 J
-850 J
-800 J
-700 J
2

In an adiabatic process, a gas expands from a volume of \( 1 \, \text{L} \) to \( 4 \, \text{L} \), reducing its pressure from \( 16 \, \text{atm} \) to \( 1 \, \text{atm} \). What is the value of \( \gamma \) (ratio of specific heats)?

For an adiabatic process, \( P_1 V_1^\gamma = P_2 V_2^\gamma \).

Substitute: \( 16 \times 1^\gamma = 1 \times 4^\gamma \).

\( 16 = 4^\gamma \).

Taking log: \( \log(16) = \gamma \log(4) \).

\( \log(16) = \log(2^4) = 4 \log(2) \), \( \log(4) = \log(2^2) = 2 \log(2) \).

\( 4 \log(2) = \gamma \times 2 \log(2) \Rightarrow \gamma = \frac{4}{2} = 2 \).

1.33
1.5
1.67
2
4

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